What is the integral of (7x^3+8x)/(x-1) ?

The integral of (7x^3+8x)/(x-1) is (7x^3)/3+(7x^2)/2+15x+15ln|x-1|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (7x^3+8x)/(x-1)

\int \frac{7 x ^{3} + 8 x}{x - 1} d x

\frac{7 x ^{3}}{3} + \frac{7 x ^{2}}{2} + 15 x + 15 ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{7 x ^{3} + 8 x}{x - 1} d x

Long division

\frac{7 x ^{3} + 8 x}{x - 1} : 7 x^{2} + 7 x + 15 + \frac{15}{x - 1}

= \int 7 x^{2} + 7 x + 15 + \frac{15}{x - 1} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 7 x^{2} d x + \int 7 x d x + \int 15 d x + \int \frac{15}{x - 1} d x

\int 7 x^{2} d x = \frac{7 x ^{3}}{3}

\int 7 x d x = \frac{7 x ^{2}}{2}

\int 15 d x = 15 x

\int \frac{15}{x - 1} d x = 15 ln ∣ x - 1 ∣

= \frac{7 x ^{3}}{3} + \frac{7 x ^{2}}{2} + 15 x + 15 ln ∣ x - 1 ∣

Add a constant to the solution

= \frac{7 x ^{3}}{3} + \frac{7 x ^{2}}{2} + 15 x + 15 ln ∣ x - 1 ∣ + C

(ln (x + x^{2} + 1))^{^{'}}

n = 0 \sum \infty 7 + e^{- 4 n}

\int \frac{3}{4} x^{2} - 5 x^{3} d x

\int sin^{3} (1) d x

n = 0 \sum \infty 2^{n} 0. 5^{n}

What is the integral of (7x^3+8x)/(x-1) ?
The integral of (7x^3+8x)/(x-1) is (7x^3)/3+(7x^2)/2+15x+15ln|x-1|+C